Relationship Between Ka and Kb: Values

Relation Between Ka, Kb, And Kw For Conjugate Pair

The relation between $\mathrm{K}_{\mathrm{a}}, \mathrm{K}_{\mathrm{b}}$ and $\mathrm{K}_{\mathrm{w}}$ can be understand by the following reaction.
$
\begin{array}{ll}
\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\
\text {Base } & \text { Conjugate acid }
\end{array}
$

The equilibrium constant $\mathrm{K}_{\mathrm{b}}$ for base $\mathrm{NH}_3$ is given as:
$
\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\mathrm{NH}_3}
$

Again, the reaction is:
$
\mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_3+\mathrm{H}_3 \mathrm{O}^{+}
$

The equilibrium constant $\mathrm{K}_{\mathrm{a}}$ for conjugate acid $\mathrm{NH}_4{ }^{+}$is given as:
$
\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{NH}_3\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\mathrm{NH}_4^{+}}
$

Thus,
$
\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}
$

Thus, $\mathrm{K}_{\mathrm{a}} \propto 1 / \mathrm{K}_{\mathrm{b}}$ at a given temperature

Therefore, if K_a of acid increases, then K_b of the conjugate base decreases. In other words, the conjugate base of a strong acid is a weak base, and vice-versa.

Buffer Solution

A buffer solution is a special type of solution that resists changes in pH when small amounts of an acid or base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Basic buffer solution

A basic buffer solution contains equimolar quantities of a weak base and its salt with strong acid. For example ammonium hydroxide i.e. NH₄OH and ammonium chloride i.e. NH₄Cl.On Adding Acid: H⁺ release and combines with OH^- of the base. On Adding Base: OH^- releases and combines with NH₄⁺ of salt. On adding acid to the basic buffer, its H⁺ ions react with OH^- ions of the base and form H₂O. Thus, in this case, the solution feels that its [OH^-] has decreased, thus to neutralize this effect, NH₄OH dissociates in small amounts and gives [OH^-] to restore the concentration of [OH^-]On adding base to the basic buffer, its [OH^-] ions react with NH₄⁺ ions and forms NH₄OH. In this case, the solution feels that its NH₄OH concentration is increased. Thus, in this case, the reaction will not proceed forward because of the common ion effect.

Recommended topic video on ( Ka and Kb Relationship)

Some Solved Examples

Example.1

1. Which one of the following is true regarding the relation between $\mathrm{k}_{\mathrm{a}}$ and $\mathrm{k}_{\mathrm{b}}$

1) $\mathrm{Pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=\mathrm{pk}_{\mathrm{w}}$
2) $\mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}}$
3) $\mathrm{Pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=14$

Solution

We know,
$
\begin{aligned}
& \mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}} \\
& \therefore \mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=\mathrm{pk}_{\mathrm{w}}=14
\end{aligned}
$

So,

All of the given equations are true regarding the relation between $\mathrm{k}_{\mathrm{a}}$ and $\mathrm{k}_{\mathrm{b}}$

Hence, the answer is the option (4).

Example.2

2. Which one of the following is equal to the $\frac{p+1}{p K_1}$ of a weak acid?

1)Its relative molecular mass

2)The $p k_b$ its conjugate base

3) (correct)The pH of a solution containing equal amount of acids and its conjugate base

4)The equilibrium concentration of its conjugate

Solution

Pk_a represents the pH of the solution when the acid is half dissociated, that is when the solution contains an equal amount of the acid and its conjugate base.

Hence, the answer is the option (3).

Example.3

3. Which statement is correct regarding the relationship between K_a and K_b of the conjugate acid-base pair?

${ }_{1)} K_w=\frac{K_b}{K_a}$
2) $K_w=\frac{K_a}{K_b}$
3) $K_w=K_a+K_b$
4) (correct) $K_w=K_a \times K_b$

Solution

Let's consider a conjugate acid-base pair HA and A^-.

The acid reacts with water as :
$
\begin{aligned}
& \mathrm{HA}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+A^{-}(a q) \\
& K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[A^{-}\right]}{[\mathrm{H} A]}
\end{aligned}
$

Conjugate base reacts with water as :
$
\begin{aligned}
& A^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HA}+\mathrm{OH}^{-} . \\
& K_b=\frac{[\mathrm{H} A]\left[\mathrm{OH}^{-}\right]}{\left[A^{-}\right]}
\end{aligned}
$

Adding equations (A) and (B)
$
\begin{gathered}
2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \\
K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]
\end{gathered}
$

By equations (1), (2), and (3)
$
K_w=K_a \times K_b
$

Hence, the answer is the option(4).

Example.4

4. The ionization constant of $N H_4^{+}$water is $6^{\star} 10^{-10}$ at 25 degrees Celcius. The rate constant for reaction of $N H_4^{+}$and $O H^{-}$to form $N H_3$ and $H_2 O$ at 25 degrees Celcius is $3.5^{\star} 10^{10}$ I/mol.sec. Calculate the rate constant for proton transfer from water to $\mathrm{NH}_3$.

1) $6.66 \times 10^3$
2) (correct) $5.83 \times 10^5$
3) $6.66 \times 10^6$
4) $7.64 \times 10^5$

Solution

The reactions according to the question will be:
$
\begin{aligned}
& \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \stackrel{K_f}{K_b} \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=3.5 \times 10^{10} \\
& \mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}^{+} ; \quad K_{\text {acid }}=6 \times 10^{-10} \\
& K_{\text {base }}=\frac{K_f}{K_b} \\
&
\end{aligned}
$

Also, $K_{\text {base }}=\frac{K_w}{K_{\text {acid }}}$
$
\begin{aligned}
& \frac{K_f}{3.5 \times 10^{10}}=\frac{10^{-14}}{6 \times 10^{-10}} \\
& K_f=5.83 \times 10^5
\end{aligned}
$

Hence, the answer is the option (2).

Example.5

5. Calculate the ionization constant of the conjugate acid of NH₃.

Given : $\mathrm{K}_{\mathrm{b}}=1.77 \times 10^{-5}$

1) $\left(\right.$ correct) $5.64 \times 10^{-10}$
2) $5.64 \times 10^{-19}$
3) $1.77 \times 10^{-10}$
4). $1.77 \times 10^{-5}$

Solution

We know that, for a conjugate acid-base pair
$
\mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}}
$

Thus,
$
\mathrm{k}_{\mathrm{a}}=\frac{\mathrm{k}_{\mathrm{w}}}{\mathrm{k}_{\mathrm{b}}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=5.64 \times 10^{-10}
$

Hence, the answer is the option (1).

6. HF and F^- represent a conjugate acid-base pair.

Which of the following is the correct relation between $\mathrm{k}_{\mathrm{a}}$ of $\mathrm{HF}_{\text {and }} \mathrm{k}_{\mathrm{b}}$ of $\mathrm{F}^{-}$at temperature T ?

(Given, at Temperature T, the self-ionization constant of Water is 10^-12)

1) $\left(\right.$ correct) $\mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=12$
2) $\mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=14$
3) $\mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=6$
4) $\mathrm{Pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=7$

Solution

As we learned,
For a conjugate acid-base pair,
$
\begin{aligned}
& \mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}} \\
& \Rightarrow \mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=\mathrm{pk}_{\mathrm{w}}=12 \\
& {\left[\because \text { At Temperature T K, } \mathrm{k}_{\mathrm{w}}=10^{-12}\right]}
\end{aligned}
$

Hence, the answer is the option (1).

Summary

Acid-base conjugate pairs are important in various chemical and biological processes. Some of the uses of conjugate acid-base pairs are Buffer Solutions in which acid-base conjugate pairs are fundamental in buffer solutions, which stops changes in pH when small amounts of acid or base are added. Ph is an important component in maintaining G the stability between the biological and the industrial process. It has several applications and one of its applications includes the regulation of pH in the components either It is biological or chemical. And the biological systems (like blood) to chemical manufacturing. For example, the bicarbonate (HCO₃⁻)/carbonic acid (H₂CO₃) pair helps maintain blood pH. Conjugate acid-base pairs are also used for chemical reactions in many reactions, conjugate pairs take part in acid-base equilibrium, affecting reaction direction and equilibrium positions.